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k0ka [10]
2 years ago
14

A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor

at 25◦C. To achieve the desired amount, we cool the tank while we fill it slowly, keeping the tank and its content at 30◦C. Find the final pressure to reach before closing the valve and the heat transfer.
Physics
1 answer:
Alex17521 [72]2 years ago
3 0

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

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A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s
Oksana_A [137]

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

5 0
2 years ago
A particle moving in the x direction is being acted upon by a net force F(x)=Cx2, for some constant C. The particle moves from x
elixir [45]

Answer:

Change in kinetic energy is ( 26CL³)/3

Explanation:

Given :

Net force applied, F(x) = Cx²  ....(1)

Displacement of the particle from xi = L to xf = 3L.

The work-energy theorem states that change in kinetic energy of the particle is equal to the net amount of work is done to displace the particle.

That is,

ΔK = W = ∫F·dx

Substitute equation (1) in the above equation.

ΔK =  ∫Cx²dx

The limit of integration from xi = L to xf = 3L, so

\Delta K=\frac{C}{3}(x_{f} ^{3} - x_{i} ^{3})

Substitute the values of xi and xf in the above equation.

\Delta K=\frac{C}{3}((3L) ^{3} - L ^{3})

\Delta K=\frac{C}{3}\times26L^{3}

5 0
2 years ago
An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o
Karo-lina-s [1.5K]

Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

4 0
2 years ago
Which of the choices describes an action–reaction force pair for a space station containing astronauts in orbit about the earth?
11111nata11111 [884]

Answer:

(D) The weight of the space station and the gravitational force of the space station on the earth.

Explanation:

In both A and B , both the forces act in the same direction ( downwards ) , so they can not be action- reaction force .

In the option C , weight of a astronaut can only be reaction force of gravitational force exerted on the earth by astronaut. Both astronaut and the earth pull each other with equal and opposite force.  So option D is correct.

5 0
2 years ago
How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that
sleet_krkn [62]

Answer: IT IS STRONGLY REPELLED

Explanation: The laws of guiding magnetic attraction or repulsion of Magnetic materials,states that when like poles are brought together they repel each other, but when unlike poles are brought together they are attracted.

The rod will be strongly repelled because the forces on the rod is greater and has the same Polarity as the charged ball.

6 0
2 years ago
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